If $p$ is an increasing polynomial function, and $c>0$ is a constant, do we always have: $$ \underset{x \rightarrow \infty}{\lim} \dfrac{p(xc+c)}{p(xc)} = 1?$$
I think if I am not missing anything, as the coefficient and degree of the leading terms will be equal in the numerator and denominator, by L'Hôspital's rule the limit should be equal to $1$. I am wondering if there is an easier proof for this fact, or anything that I am missing.
On
Yes, there's an easier proof. Divide the numerator and denominator of the fraction by $x^n$, where $n = \deg p$. Then your numerator and denominator both end up as $a_nc^n$ (where $a_n$ is the coefficient of the $x^n$ term of your polynomial) plus a bunch of terms that go to $0$ as $x \to \infty$. By the definition of degree, $a_n \neq 0$, and you're given that $c \neq 0$ (in fact, $c \gt 0$), so this fraction goes to $1$.
On
It is true what you say. Recall that if $P(x)$ and $Q(x)$ are two polynomials of the same degree, then $\displaystyle \lim_{x\to \pm \infty}\dfrac{P(x)}{Q(x)}=\dfrac{a_n}{b_n}$, with $a_n$ and $b_n$ being the leading coefficients of $P(x)$ and $Q(x)$, respectively.
In the case we are considering, the two polynomials have the same degree and $a_n=kc^n=b_n$, where $k$ is the principal coefficient of $p$. Therefore,
$$\lim_{x\to \infty}\dfrac{p(xc+c)}{p(xc)}=\dfrac{kc^n}{kc^n}=1.$$
On
A polynomial can be written as: $$a(x-\lambda_1)(x-\lambda_2)\dots (x-\lambda_n)$$ where $\lambda_1,\lambda_2\dots\lambda_n \in \mathbb{C}$
The leading coeficient $a$ factors out of the ratio, which "distributes" over the sub-polynomials, so it is sufficient to prove $$\lim_{x\to \infty}\dfrac{p(xc+c)}{p(xc)}=1$$ for $p(x) = (x-\lambda_{k})$ . We get $\frac{xc + c-\lambda_{k}}{xc-\lambda_{k}} = 1 + \frac{c}{xc-\lambda_{k}} \to 1$ as $x \to \infty$.
You don't need L'Hopital's Rule to prove this. If$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$with $N\in\Bbb N$ and $a_n\ne0$, then\begin{align}\lim_{x\to\infty}\frac{p(xc+c)}{p(xc)}&=\lim_{x\to\infty}\frac{p\bigl((x+1)c\bigr)}{p(xc)}\\&=\lim_{x\to\infty}\frac{a_n\bigl((x+1)c\bigr)^n+a_{n-1}\bigl((x+1)c\bigr)^{n-1}+\cdots+a_1\bigl((x+1)c\bigr)+a_0}{a_n(xc)^n+a_{n-1}(xc)^{n-1}+\cdots+a_1(xc)+a_0}\\&=\lim_{x\to\infty}\frac{a_nc^nx^n+\text{terms of smaller degree}}{a_nc^nx^n+\text{terms of smaller degree}}\end{align}and this limit is equal to $1$; just divide both the numerator and the denominator by $x^n$. Then it's clear that the limit is $\frac{a_nc^n}{a_nc^n}=1$.